25-Nav-A1 Fundamentals of Naval Architecture · May-98-Mar-A1 2017

Question 7 of 8: Dewar Vessel with Radiation Shield

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Examinations, May 2017 — 98-Mar-A1 Applied Thermodynamics and Heat Transfer, 3 hours, open book (Part A: Thermodynamics, Part B: Heat Transfer; 5 of 8 questions required, all 8 answered below for full study coverage).

Reference texts: Cengel & Boles, Thermodynamics: An Engineering Approach; Sonntag, Borgnakke & Van Wylen, Fundamentals of Thermodynamics; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer.

It is solved as the thermodynamics/heat-transfer exam it actually is.

Question 7: Dewar Vessel with Radiation Shield (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Given data
Quantity	Symbol	Value
Inner s‌hell diameter, temperature	$D_1,T_1$	510 mm, 90 K (LOX)
Radiation-shield diameter	$D_s$	560 mm
Outer s‌hell diameter, temperature	$D_2,T_2$	610 mm, 45°C (318.15 K)
Emissivity, all surfaces	$\varepsilon$	0.3

Concentric-sphere Dewar with a single floating radiation shield.

Find. Rate of heat gain to the liquid oxygen.

Approach. Model the evacuated (vacuum, no conduction/convection) gap as two concentric-sphere radiation resistances in series — inner s‌hell to shield, then shield to outer s‌hell — and sum them before applying the Stefan–Boltzmann driving potential $T_2^4-T_1^4$.

Surface areas. $A_1=4\pi(0.255)^2=0.8171\text{ m}^2$, $A_s=4\pi(0.28)^2=0.9852\text{ m}^2$, $A_2=4\pi(0.305)^2=1.169\text{ m}^2$.
Resistance, inner s‌hell↔shield. $$\dfrac{1}{A_1}\left[\dfrac1\varepsilon+\dfrac{A_1}{A_s}\left(\dfrac1\varepsilon-1\right)\right]=\dfrac{1}{0.8171}\left[3.333+0.829(2.333)\right]=6.448\text{ m}^{-2}$$
Resistance, shield↔outer s‌hell. $$\dfrac{1}{A_s}\left[\dfrac1\varepsilon+\dfrac{A_s}{A_2}\left(\dfrac1\varepsilon-1\right)\right]=\dfrac{1}{0.9852}\left[3.333+0.843(2.333)\right]=5.379\text{ m}^{-2}$$
Heat gain to the LOX. $R_{tot}=6.448+5.379=11.83\text{ m}^{-2}$, so $$\boxed{\dot Q=\dfrac{\sigma(T_2^4-T_1^4)}{R_{tot}}=\dfrac{5.67\times10^{-8}(318.15^4-90^4)}{11.83}=48.8\text{ W}}$$ (for reference, with no shield at all the same driving potential gives $\approx95.0\text{ W}$ — the single shield roughly halves the heat gain.)

Question 7 — final results
Quantity	Value
Heat gain to liquid oxygen (with shield)	$\dot Q = 48.8\text{ W}$
Heat gain, no shield (reference)	$\dot Q \approx 95.0\text{ W}$

Topic: Radiation exchange between diffuse-gray concentric spheres with a radiation shield.

Key relations: two-surface enclosure $\dot Q_{12}=\sigma A_1(T_1^4-T_2^4)/\left[\tfrac1{\varepsilon_1}+\tfrac{A_1}{A_2}\left(\tfrac1{\varepsilon_2}-1\right)\right]$; a floating shield adds its own resistance term in series, with area ratios evaluated at each interface.

Why this works: A radiation shield intercepts and re-radiates rather than blocking heat entirely; because it floats at an equilibrium temperature between the two bounding surfaces, its two resistance terms simply add in series to the original single-surface-pair resistance, exactly like conduction resistances in series.

Common pitfall: Using the inner sphere's area $A_1$ throughout instead of the correct area at each radiation interface ($A_1/A_s$ for the inner leg, $A_s/A_2$ for the outer leg) — the areas differ enough here (0.82, 0.99, 1.17 m²) that this materially changes the split of the total resistance.

Source: Incropera & DeWitt, Fundamentals of Heat and Mass Transfer, Ch. 13 (radiation exchange between surfaces — radiation shields).

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